Atomic exitations, difference between photon and electron collision?

[kof9595995]

I guess this is a FAQ, but since it constantly confused me, I'm just going to ask anyway.
Two ways of exciting atoms, shoot a beam of photons and let atoms absorb them, or shoot a beam of electrons, and let atoms collide with them.
However, for the excitations to happen, the former case requires the photon energy matches the energy level differences exactly, and the latter case only requires kinetic energy of electrons bigger then energy difference. How should we explain this from first principles of QM or QFT?

 

[Drakkith]

Hrmm. I'll give this a shot. Hopefully I'm mostly correct here. =)

As far as I know, a Photon can only be absorbed as a whole, or not at all. If it has too much energy for 1 level, but not enough for 2, then it can't be absorbed as there is nowhere for the extra energy to go. If it has too little energy then again there is nowhere for the energy to go. Since photons are massless and always move the same speed, they don't have kinetic energy and instead have only momentum.

For electrons, they have mass and kinetic energy, not merely momentum as photons do. The electrons can impart their kinetic energy in just about any amount to the whole atom instead of just the electron energy levels. An electron hitting an atom can impart its energy to the whole atom in the form of movement of the atom.

 

[Isaacsname]

Hrmm. I'll give this a shot. Hopefully I'm mostly correct here. =)

As far as I know, a Photon can only be absorbed as a whole, or not at all. If it has too much energy for 1 level, but not enough for 2, then it can't be absorbed as there is nowhere for the extra energy to go. If it has too little energy then again there is nowhere for the energy to go. Since photons are massless and always move the same speed, they don't have kinetic energy and instead have only momentum.

For electrons, they have mass and kinetic energy, not merely momentum as photons do. The electrons can impart their kinetic energy in just about any amount to the whole atom instead of just the electron energy levels. An electron hitting an atom can impart its energy to the whole atom in the form of movement of the atom.

Drakkith, please excuse my ignorance here, I keep reading photons are massless, but that radiation has pressure ? I know it's supposed to be very weak, but is this disregarded ?

Thanks

 

[Drakkith]

Drakkith, please excuse my ignorance here, I keep reading photons are massless, but that radiation has pressure ? I know it's supposed to be very weak, but is this disregarded ?

Thanks

The pressure is from the momentum of the photon. It is massless, but it does posess momentum. Its energy is related to it's frequency, and since it travels at c, you can calculate the momentum.

See here. http://en.wikipedia.org/wiki/Photon#Physical_properties

 

[Isaacsname]

The pressure is from the momentum of the photon. It is massless, but it does posess momentum. Its energy is related to it's frequency, and since it travels at c, you can calculate the momentum.

See here. http://en.wikipedia.org/wiki/Photon#Physical_properties

Thank you.

 

[kof9595995]

If it has too much energy for 1 level, but not enough for 2, then it can't be absorbed as there is nowhere for the extra energy to go. If it has too little energy then again there is nowhere for the energy to go. Since photons are massless and always move the same speed, they don't have kinetic energy and instead have only momentum.

Why, I thought photons possesses only kinetic energy because they are massless. And why can't the atom absorb the photon and just emit another lower frequency photon at the same time?

 

[ZapperZ]

I guess this is a FAQ, but since it constantly confused me, I'm just going to ask anyway.
Two ways of exciting atoms, shoot a beam of photons and let atoms absorb them, or shoot a beam of electrons, and let atoms collide with them.
However, for the excitations to happen, the former case requires the photon energy matches the energy level differences exactly, and the latter case only requires kinetic energy of electrons bigger then energy difference. How should we explain this from first principles of QM or QFT?

So in this case, you are ruling out the scenario for complete ionization? Because in that case, there really isn't that restriction for photons as well. The extra energy is carried away predominantly by the liberated electron.

Zz.

 

[alxm]

You also have inelastic (Raman) and elastic (Rayleigh) scattering of photons off atoms and molecules.

 

[dr_uri]

The following points might be of use:
1. Photon impact excitation cross sections are calculated with Fermi's Golden rule (see Sakurai Advanced Quantum Mechanics p39-57 onwards). The calculation leads to a delta function selection rule: E_f = E_i + \hbar\omega.
2. Electron impact excitation cross sections can be obtained with the aid of the Lippmann-Schwinger equation (read Roman, Advanced Quantum Theory and learn about the link between the T matrix, S matrix and differential cross section). The critical point is that the V matrix elements in the Lippmann-Schwinger equation do not have the delta function selection rule involving \hbar\omega.

Put simply, the interaction terms in photon-atom collisions differ from the interaction terms in electron-atom collisions. See eq 2.102 Sakurai for the former. The latter involve the coulomb interation (1/r) + any relativistic effects (eg Breit interaction) of relevance.

Hope that helps; a full understanding will require a lot of reading.

 

[kof9595995]

So in this case, you are ruling out the scenario for complete ionization? Because in that case, there really isn't that restriction for photons as well. The extra energy is carried away predominantly by the liberated electron.
.

Yes, I'm only talking about bound state transition

 

[kof9595995]

The following points might be of use:
1. Photon impact excitation cross sections are calculated with Fermi's Golden rule (see Sakurai Advanced Quantum Mechanics p39-57 onwards). The calculation leads to a delta function selection rule: E_f = E_i + \hbar\omega.

But that delta function results from large t limit, which means for small t, non-resonant frequency can also induce a transition.

 

[dr_uri]

kof, please read the references carefully, particularly Sakurai p57.

Yes the delta function does result from the large t limit. Standard Fourier analysis mandates that a well defined frequency in the frequency domain corresponds to an infinite time extent in the time domain. If we have smaller t, then we no longer have a well defined photon frequency, and your question related to a particular "photon energy" cannot be addressed.

 

[kof9595995]

Emm, I can't fully understand it at the time because I'm not familiar with some technical terms he used for the discussion, but I see it is at least highly related to my question, thanks.

 

[kof9595995]

Standard Fourier analysis mandates that a well defined frequency in the frequency domain corresponds to an infinite time extent in the time domain.

Actually, I'd like to know how exactly Fourier analysis results in this. I see this conclusion so often that I take it for granted, partially because it reminds me of uncertainty principle. But there is no energy-time uncertainty principle in any common sense. So what is the exact meaning of what I just quoted?

 

[kof9595995]

Why, I thought photons possesses only kinetic energy because they are massless. And why can't the atom absorb the photon and just emit another lower frequency photon at the same time?

As I just read from sakurai, this process is indeed possible(but absorption and emission need not to be simultaneous) and is responsible for the so called Stokes' line, a spectral line more reddish than that of incident radiation.